Ochem Destroyer 2011 # 119

[aliss4]

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I remember Chad mentioned that only with TERTIARY alkyl halides and bulky base, we will get hoffman product( anti-zaitsev), but here we also got hoffman product, even though our alkyl halide was secondary. anyone?😕


It's destroyer 2010 ( thanks to PooaH)

 

[PooyaH]

Are you sure it's #119 you're talking about?!

 

[budda10000]

i beleive tert. substrate will always favor e2, w/ a smaller base it can pull a proton off a harder to access point but with bulky base it will go for the most available.

 

[aliss4]

Are you sure it's #119 you're talking about?!

yes. The reactant is 2-butanol which reacts with 1) SOCL2 then 2)
(CH3)3CO-K+

 

[mh0000]

I'm not sure if I agree that ONLY tertiary halides with a bulky base will undergo hoffman elimination.You might get a mixture of products with a secondary. Are there any other bulky groups around the halide?

 

[aliss4]

i beleive tert. substrate will always favor e2, w/ a smaller base it can pull a proton off a harder to access point but with bulky base it will go for the most available.

That's right but these are the exact words of chad " some teachers say, every time you use a bulky base, you get hoffman, NOT TRUE. Technically if you have a tertiary alkyl halide and bulky base, you will get hoffman product."

 

[PooyaH]

yes. The reactant is 2-butanol which reacts with 1) SOCL2 then 2)
(CH3)3CO-K+

I have the 2011 version too and #119 on here doesn't even have a reaction, it's a wordy question! I'm not hallucinating am I? lol

 

[aliss4]

I'm not sure if I agree that ONLY tertiary halides with a bulky base will undergo hoffman elimination.You might get a mixture of products with a secondary. Are there any other bulky groups around the halide?

No it's only 2-butanol with a bulky base

 

[aliss4]

I have the 2011 version too and #119 on here doesn't even have a reaction, it's a wordy question! I'm not hallucinating am I? lol

oh sorry it's 2010😀. my bad

 

[PooyaH]

oh sorry it's 2010😀. my bad

Thank god, I was gonna stop studying immediately! haha

 

[aliss4]

That's right but these are the exact words of chad " some teachers say, every time you use a bulky base, you get hoffman, NOT TRUE. Technically if you have a tertiary alkyl halide and bulky base, you will get hoffman product."

he also added, " if your halide is a secondary, you don't always get hoffman, you often get the zait's product." Yes I memorized all of his notes, haha.

 

[PooyaH]

he also added, " if your halide is a secondary, you don't always get hoffman, you often get the zait's product." Yes I memorized all of his notes, haha.

I'm looking at a problem in 'Organic Chemistry Odyssey' which is also written by Romano, and it says secondary halides only show Hoffman products when reacted with strong bulky bases!

2-bromo-pentane is reacted with (CH3)3CO- & you get pentene.

 

[aliss4]

I'm looking at a problem in 'Organic Chemistry Odyssey' which is also written by Romano, and it says secondary halides only show Hoffman products when reacted with strong bulky bases!

2-bromo-pentane is reacted with (CH3)3CO- & you get pentene.

so chad made a mistake??? so tertiary and secondary give us hoffman, primary gives us zaits product. Right?

 

[PooyaH]

so chad made a mistake??? so tertiary and secondary give us hoffman, primary gives us zaits product. Right?

Well primary halides don't go through elimination, It's almost always SN2. Secondary & tertiary halides go through elimination and give us Hoffman products with a strong bulky base.

I lied! primary halides do go through elimination only with strong bulky bases! Got that mixed up! Sorry.

 

[aliss4]

Well primary halides don't go through elimination, It's almost always SN2. Secondary & tertiary halides go through elimination and give us Hoffman products with a strong bulky base.

actually even in the case of primary alkyl halides, elimination is favored when we use a bulky base. but in respond to my own question, primary alkyl halides has only one choice and will gives us only 1 product. It's secondary and tertiary that can give us both hoffman and zaits.

 

[PooyaH]

actually even in the case of primary alkyl halides, elimination is favored when we use a bulky base. but in respond to my own question, primary alkyl halides has only one choice and will gives us only 1 product. It's secondary and tertiary that can give us both hoffman and zaits.

Yes you're right, I did correct myself though. But secondary and tertiary halides always make Hoffman products with strong bulky bases.

 

[aliss4]

actually even in the case of primary alkyl halides, elimination is favored when we use a bulky base. but in respond to my own question, primary alkyl halides has only one choice and will gives us only 1 product. It's secondary and tertiary that can give us both hoffman and zaits.

Oh I found this link:

http://forums.studentdoctor.net/showthread.php?p=10652002

Apparently we are not the only confused people here lol

 

[aliss4]

Yes you're right, I did correct myself though. But secondary and tertiary halides always make Hoffman products with strong bulky bases.

Yes🙂 thanks PooyaH ! now go study😀

 

[PooyaH]

Yes🙂 thanks PooyaH ! now go study😀

Indeed. It seems like I need to! lol

 

[rpatel8]

I think for DAT purposes, we can just say that a bulky strong base gives the less substituted alkene.